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Aleksandr [31]
3 years ago
13

How many grams of oxygen are required to react with 12.0 grams of octane in the combustion of octane in gasoline?

Chemistry
1 answer:
Svetach [21]3 years ago
6 0
2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) 

<span>(10.0 g C8H18) / (114.2285 g C8H18/mol) x (25 mol O2 / 2 mol C8H18) x (31.99886 g O2/mol) = 35.0 g O2
please return the favor and answer question in my profile thanks 
</span>
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NaOH (aq) + HCl(aq) H2O + NaCl (aq)
yuradex [85]

Answer:

The answer to your question is 0.005

Explanation:

Data

Volume of NaOH = 25 ml

[NaOH] = 0.2 M

moles of NaOH = ?

To solve this problem is not necessary to have the chemical reaction. Just use the formula of Molarity and solve it for moles.

Formula

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Convert volume to liters

          1000 ml ---------------- 1 l

              25 ml ---------------- x

               x = (25 x 1) / 1000

               x = 0.025 l

-Substitution

moles = 0.2 x 0.025

-Result

moles = 0.005

7 0
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Answer:

0.65moles of HCl was produced

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Explanation:

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